The entropy change of the computer chips.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

36 2) Use the method of MEMBERS to determine the true magnitude and direction of the forces in members1 and 2 of the frame shown below in Fig 3.2. 300lbs/ft member-1 member-2 30° Fig 3.2. https://brightspace.cuny.edu/d21/le/content/433117/viewContent/29873977/View

Can you solve this for me?

5670 mm The apartment in the ground floor of three floors building in Fig. in Baghdad city. The details of walls, roof, windows and door are shown. The window is a double glazing and air space thickness is 1.3cm Poorly Fitted-with Storm Sash with wood strip and storm window of 0.6 cm glass thickness. The thickness of door is 2.5 cm. The door is Poor Installation. There are two peoples in each room. The height of room is 280 cm. assume the indoor design conditions are 25°C DBT and 50 RH, and moisture content of 8 gw/kga. The moisture content of outdoor is 10.5 gw/kga. Calculate heat gain for living room : الشقة في الطابق الأرضي من مبنى ثلاثة طوابق في مدينة بغداد يظهر في مخطط الشقة تفاصيل الجدران والسقف والنوافذ والباب. النافذة عبارة عن زجاج مزدوج وسمك الفراغ الهوائي 1.3 سم ضعيف الاحكام مع ساتر حماية مع إطار خشبي والنافذة بسماكة زجاج 0.6 سم سماكة الباب 2.5 سم. الباب هو تركيب ضعيف هناك شخصان في كل غرفة. ارتفاع الغرفة 280 سم. افترض أن ظروف التصميم الداخلي هي DBT25 و R50 ، ومحتوى الرطوبة 8…

Answer 2

Question

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Answer 6

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Answer 7

Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

Answer 8

(a)

Expert Solution

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Answer 13

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Answer 14

(b)

To determine

The entropy change of the R-134.

Answer 15

(b)

Expert Solution

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Answer 20

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Answer 21

(c)

To determine

The entropy change of the entire system.

Answer 22

(c)

Expert Solution

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

Answer 27

Ch. 7.13 - Prob. 1P Ch. 7.13 - Does the cyclic integral of heat have to be zero...Ch. 7.13 - Is a quantity whose cyclic integral is zero...Ch. 7.13 - Prob. 4P Ch. 7.13 - Prob. 5P Ch. 7.13 - How do the values of the integral 12Q/T compare...Ch. 7.13 - The entropy of a hot baked potato decreases as it...Ch. 7.13 - When a system is adiabatic, what can be said about...Ch. 7.13 - Prob. 9P Ch. 7.13 - A pistoncylinder device contains helium gas....

The entropy change of the computer chips.

Concept explainers

Videos

The entropy change of the computer chips.

Answer to Problem 61P

Explanation of Solution

The entropy change of the R-134.

Answer to Problem 61P

Explanation of Solution

The entropy change of the entire system.

Answer to Problem 61P

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions